cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A219266 Logarithmic derivative of the superfactorials (A000178).

Original entry on oeis.org

1, 3, 31, 1103, 171311, 149089887, 877704854447, 40451674467223423, 16514355739866259408591, 66586047491662065505372477983, 2923692867015618804999172694908629103, 1527767556403309713534536695030930443376591295, 10306227067090276816548435451550663056418226402352755215
Offset: 1

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Author

Paul D. Hanna, Nov 16 2012

Keywords

Comments

Superfactorial A000178(n) equals the product of first n factorials.

Examples

			L.g.f.: L(x) = x + 3*x^2/2 + 31*x^3/3 + 1103*x^4/4 + 171311*x^5/5 +...
where
exp(L(x)) = 1 + x + 2*x^2 + 12*x^3 + 288*x^4 + 34560*x^5 + 24883200*x^6 + 125411328000*x^7 +...+ n!*(n-1)!*(n-2)!*...*3!*2!*1!*0!**x^n +...
		

Crossrefs

Programs

  • Mathematica
    nmax=15; Rest[CoefficientList[Series[Log[Sum[BarnesG[k+2]*x^k,{k,0,nmax}]],{x,0,nmax}],x] * Range[0,nmax]] (* Vaclav Kotesovec, Jul 10 2015 *)
  • PARI
    {a(n)=n*polcoeff(log(sum(k=0,n+1,prod(j=0,k,j!)*x^k)+x*O(x^n)),n)}
    for(n=1,21,print1(a(n),", "))

Formula

a(n) ~ n^(n^2/2 + n + 17/12) * (2*Pi)^((n+1)/2) / (A * exp(3*n^2/4 + n - 1/12)), where A = A074962 = 1.2824271291... is the Glaisher-Kinkelin constant. - Vaclav Kotesovec, Jul 10 2015

A219270 G.f. satisfies: A(x) = Sum_{n>=0} x^n * A(x)^n * Product_{k=0..n} k!.

Original entry on oeis.org

1, 1, 3, 19, 357, 36301, 25099919, 125586786039, 5057589140280713, 1834978988294513263993, 6658624933768189847669906875, 265790340541205664455435665865706651, 127313966488883223582008076686428073754622381, 792786699250878346141762323574616271109902136066501317
Offset: 0

Views

Author

Paul D. Hanna, Nov 16 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 19*x^3 + 357*x^4 + 36301*x^5 + 25099919*x^6 +...
where
A(x) = 1 + 1!*x*A(x) + 2!*1!*A(x)^2 + 3!*2!*1!*A(x)^3 + 4!*3!*2!*1!*A(x)^4 +...
Given F(x) is the g.f. of the superfactorials A000178:
F(x) = 1 + x + 2*x^2 + 12*x^3 + 288*x^4 + 34560*x^5 + 24883200*x^6 +...
then the table of coefficients in F(x)^n begins:
n=1: [(1), 1, 2, 12, 288, 34560, 24883200, 125411328000, ...];
n=2: [1,(2), 5, 28, 604, 69744, 49836816, 250872567552, ...];
n=3: [1, 3, (9), 49, 954, 105600, 74861864, 376383826368, ...];
n=4: [1, 4, 14, (76), 1345, 142184, 99959432, 501945213632, ...];
n=5: [1, 5, 20, 110,(1785), 179561, 125130690, 627556840100, ...];
n=6: [1, 6, 27, 152, 2283,(217806), 150376901, 753218818212, ...];
n=7: [1, 7, 35, 203, 2849, 257005,(175699433), 878931262217, ...];
n=8: [1, 8, 44, 264, 3494, 297256, 201099772,(1004694288312), ...]; ...
in which the main diagonal generates this sequence:
[1/1, 2/2, 9/3, 76/4, 1785/5, 217806/6, 175699433/7, 1004694288312/8, ...].
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(1/x*serreverse(x/sum(k=0,n+1,prod(j=0,k,j!)*x^k)+x^2*O(x^n)),n)}
    for(n=0,21,print1(a(n),", "))

Formula

a(n) = [x^n] F(x)^(n+1)/(n+1) where F(x) = Sum_{n>=0} x^n*Product_{k=0..n} k! is the g.f. of the superfactorials A000178.
G.f.: (1/x)*Series_Reversion(x/F(x)) where F(x) = Sum_{n>=0} x^n*Product_{k=0..n} k! is the g.f. of A000178.
Showing 1-2 of 2 results.